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...= 0; for( int k = 0; k < 2*N; k++ ) { sum += in[k]*cos(M_PI*(2*j+1+N)*(2*k+1)/8/N); } out[j] = sum; } } > It just a normal shifted mdct, like the one used in mpeg layer 3. > it's equations are: > (time domain: a_j, freq domain: b_k; w_j is the window) > forward: > b_k = sum(j=0..4N-1) a_j * w_j * cos( (2k+1)(2j+1+-N) / 8N ) // forward > a_j = w_j * sum(k=0..2N-1) a_j * cos( (2k+1)(2j+1+-N) / 8N ) // backward (missing PI in the cos, and a_j should be b_k in summation) > where +- means: I'm not sure whether it is +...

Hi, Can someone tell me which MDCT and invMDCT equation uses? I implemented the invMDCT one given in eusipco.corrected.ps file (handed out by Monty way back) and it produces different time domain samples. I tried both the FFT method and the slow way directly from the equation and couldn't reproduce the results from the original code. This leads me to believe that the forward MDCT used in

...results here and I think I am missing something. I am trying to program a function that for a set of random variables drawn from uniform distributions plots that distribution of the second order statistic of the ordered variables. (ie I have n uniform distributions on [0, w_i] for w_i different w_j and i=1..n. I want to plot the distribution of the second order statistic ie one less the maximum. I thought that the way to do this is to calculate: F= Sum over i { (1-Fi) * Product of all j different i of Fj} + Product over all i of Fi where Fi are just the respective uniform cdf for variab...

...nd besides, I don't have access to TeX, X, or anything else from a text terminal with gcc and vi right now). Hope you like it, Segher -- Suppose we want to calculate a MDCT (encode side): (x_ is input (time) samples, w_ is window, b_ is output (freq) samples): b_k = \sum_{j=0}^{4N-1} x_j w_j \cos{(2k+1)(2j+1-2N) \over 8N} \pi After setting a_j := w_{-j-1+N} x{-j-1+N} + w_{j+N} x{j+N}, j < N a_j := w_{j-1-N} x{j-1-N} - w_{j+3N} x{j+3N}, otherwise (or something similar; the cos'es should agree) we get b_k = \sum_{j=0}^{2N-1} a_j \cos{(2k+1)(2j+1) \over 8N} \pi This is a...

...e, that would set p=0 & q=1 for Fleming-Harrington weights. > > My question is how I can do the same by setting certain values for "rho" in the regular survival() function? I think that survdiff uses a different version than what you have found. The G-rho family weights are: w_j = [S?(tj)]^? So rather than two parameters on S(t) and (1-S(t)) as in the p,q version, you only have one parameter applied to S(t). This class handout says that the G-rho,gamma weighting scheme is not available in survdiff. -- David Winsemius Alameda, CA, USA 'Any technology distinguishable...

...q=1 for Fleming-Harrington weights. >> >> My question is how I can do the same by setting certain values for "rho" in the regular survival() function? > > I think that survdiff uses a different version than what you have found. The G-rho family weights are: > > w_j = [S?(tj)]^? > > So rather than two parameters on S(t) and (1-S(t)) as in the p,q version, you only have one parameter applied to S(t). This class handout says that the G-rho,gamma weighting scheme is not available in survdiff. > Forgot to paste the link: http://www.ics.uci.edu/~dgille...

Hi all,? The survdiff() from survival package has an argument "rho" that implements Fleming-Harrington weighted long rank test.? But according to several sources including "survminer" package (https://cran.r-project.org/web/packages/survminer/vignettes/Specifiying_weights_in_log-rank_comparisons.html), Fleming-Harrington weighted log-rank test should have 2 parameters

Dear R developers, I am a little disappointed that my bug report only made it to the wishlist, with the argument: Well, it does not say it has. Only relevant to prediction intervals. predict.lm does calculate prediction intervals for linear models from weighted regression, so they should be correct, right? As far as I can see they are bound to be wrong in almost all cases, if no weights